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In $\Bbb{R}^n$, how is $L_X(dx^i)=dL_X(x^i)$? This is a calculation that is given on pg 23 of Peter Petersen's book "Riemannian Geometry".

Is this because $(dL_X)(x^i)=0$, and hence $d(L_X(x^i))=L_X(dx^i)$? I know that this is true for $\nabla$, but don't know if this is also true for the operator $d$.

  • What exactly do you not understand of the calculation? – Gibbs Feb 28 '19 at 22:05
  • @Gibbs- There seems to be no calculation here. It's just an assertion, that I have not seen before. Why is $dL_X(x^i)=L_X(dx^i)$? –  Feb 28 '19 at 22:07

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You can do that computation by hand. If $\varphi_t$ is the one-parameter group of local transformations on $\mathbb{R}^n$, then by definition $$(\mathcal{L}_X\mathrm{d}x^i)_p = \lim_{t\to 0}\frac{(\varphi_t)^*_{\varphi_t(p)}\mathrm{d}x^i_{\varphi_t(p)}-\mathrm{d}x^i_{p}}{t}. $$ Since $\varphi_t^*$ and $\mathrm{d}$ commute you have $$\lim_{t\to 0}\frac{\mathrm{d}\left((\varphi_t)^*_{\varphi_t(p)}x^i_{\varphi_t(p)}\right)-\mathrm{d}x^i_{p}}{t} = \mathrm{d}\left(\lim_{t \to 0}\frac{(\varphi_t)^*_{\varphi_t(p)}x^i_{\varphi_t(p)}-x^i_{p}}{t}\right)=\mathrm{d}(\mathcal{L}_Xx^i)_p.$$

Gibbs
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